description/proof of that for group with topology with continuous operations (especially, topological group), neighborhood of \(1\), and positive natural number, there is symmetric neighborhood of \(1\) whose power to natural number is contained in neighborhood
Topics
About: group
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of group.
- The reader knows a definition of topological space.
- The reader knows a definition of continuous map.
- The reader knows a definition of neighborhood of point on topological space.
- The reader knows a definition of symmetric subset of group.
- The reader admits the proposition that for any group with any topology with any continuous operations (especially, topological group), any finite multiplication map is continuous.
- The reader admits the proposition that for any group with any topology with any continuous operations (especially, topological group), the set of the symmetric neighborhoods of \(1\) is a neighborhood basis at \(1\).
Target Context
- The reader will have a description and a proof of the proposition that for any group with any topology with any continuous operations (especially, topological group), any neighborhood of \(1\), and any positive natural number, there is a symmetric neighborhood of \(1\) whose power to the natural number is contained in the neighborhood.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\(G\): \(\in \{\text{ the groups }\}\) with any topology such that the group operations are continuous
\(N'_1\): \(\in \{\text{ the neighborhoods of } 1 \text{ on } G\}\)
\(n\): \(\in \mathbb{N} \setminus \{0\}\)
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Statements:
\(\exists N_1 \in \{\text{ the symmetric neighborhoods of } 1 \text{ on } G\} (N_1^n \subseteq N'_1)\)
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2: Note
We do not have any immediate reason why \(G\) should not be Hausdorff (so, \(G\) should not be any topological group), but we use this proposition for proving that a \(G\) is Hausdorff, so, Hausdorff-ness is not presupposed.
3: Proof
Whole Strategy: Step 1: think of the continuous \(n\)-multiplication map, \(f_n: G \times ... \times G \to G\), and take an open neighborhood of \(1\), \(U_1\), such that \(f_n (U_1 \times ... \times U_1) \subseteq N'_1\); Step 2: take a symmetric neighborhood of \(1\), \(N_1\), such that \(N_1 \subseteq U_1\), and see that \(N_1^n = f (N_1 \times ... \times N_1) \subseteq N'_1\).
Step 1:
Let us think of the \(n\)-multiplication map, \(f_n: G \times ... \times G \to G, (g_1, ..., g_n) \mapsto g_1 ... g_n\).
\(f_n\) is continuous, by the proposition that for any group with any topology with any continuous operations (especially, topological group), any finite multiplication map is continuous.
So, as \(f_n (1, ..., 1) = 1\), there is an open neighborhood of \((1, ..., 1) \in G \times ... \times G\), \(U_{(1, ..., 1)} \subseteq G \times ... \times G\), such that \(f_n (U_{(1, ..., 1)}) \subseteq N'_1\).
There is an open neighborhood of \(1\), \(U_1 \subseteq G\), such that \((1, ..., 1) \in U_1 \times ... \times U_1 \subseteq U_{(1, ..., 1)}\), by the definition of product topology: while there are some open neighborhoods of \(1\), \(U_{1, 1}, ... U_{1, n} \subseteq G\), such that \((1, ..., 1) \in U_{1, 1} \times ... \times U_{1, n} \subseteq U_{(1, ..., 1)}\), we can take \(U_1 := U_{1, 1} \cap ... \cap U_{1, n}\).
Step 2:
There is a symmetric neighborhood of \(1\), \(N_1\), such that \(N_1 \subseteq U_1\), by the proposition that for any group with any topology with any continuous operations (especially, topological group), the set of the symmetric neighborhoods of \(1\) is a neighborhood basis at \(1\).
\(f_n (N_1 \times ... \times N_1) \subseteq f_n (U_1 \times ... \times U_1) \subseteq f_n (U_{(1, ..., 1)}) \subseteq N'_1\).
Let us see that \(f_n (N_1 \times ... N_1) = N_1^n\).
For each \(g \in f_n (N_1 \times ... \times N_1)\), \(g = f_n ((g_1, ..., g_n)) = g_1 ... g_n \in N_1^n\); for each \(g \in N_1^n\), \(g = g_1 ... g_n = f_n ((g_1, ..., g_n)) \in f_n (N_1 \times ... \times N_1)\).
So, \(N_1^n = f (N_1 \times ... \times N_1) \subseteq N'_1\).
